In a series LC circuit, L = 100 µH, C = 0.047 µF, and f = 150 kHz. The value of ZT is:
A) 72 $\Omega \angle$
90°
B) 36 $\Omega \angle$
45°
C) -72 $\Omega \angle$
-90°
D) 72 $\Omega \angle$
-90°

Question

In a series LC circuit, L = 100 µH, C = 0.047 µF, and f = 150 kHz. The value of ZT is:
A) 72 $\Omega \angle$
 90°
B) 36 $\Omega \angle$
 45°
C) -72 $\Omega \angle$
 -90°
D) 72 $\Omega \angle$
 -90°

Quizplus · Accepted Answer

The impedance of a series LC circuit is given by:ZT = R + j(XL - XC)where:R is the resistance (in ohms)XL is the inductive reactance (in ohms)XC is the capacitive reactance (in ohms)In this case, we have:L = 100 µH = 0.1 mHC = 0.047 µF = 47 nFf = 150 kHzThe inductive reactance is given by:XL = 2πfL = 2π(150 kHz)(0.1 mH) = 94.25 ohmsThe capacitive reactance is given by:XC = 1/(2πfC) = 1/(2π(150 kHz)(47 nF)) = 21.5 ohmsSince the circuit is a series LC circuit, the resistance is zero. Therefore, the impedance is:ZT = j(XL - XC) = j(94.25 ohms - 21.5 ohms) = j72.75 ohmsThe magnitude of the impedance is:|ZT| = √(Re(ZT)² + Im(ZT)²) = √(0² + 72.75²) = 72.75 ohmsThe angle of the impedance is:∠ZT = arctan(Im(ZT)/Re(ZT)) = arctan(72.75/0) = 90°Therefore, the value of ZT is 72.75 ohms at an angle of 90°.

In a Series LC Circuit, L = 100 µH, C Ω∠\Omega \angleΩ∠

In a Series LC Circuit, L = 100 µH, C $\Omega \angle$